Q: A nature photographer wishes to photograph a $28\text{ m}$ tall tree from a distance of $58\text{ m}$. What focal length of lens should be used if the image is to fill the $24\text{ mm}$ height of the film?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $OO' = +28\text{ m}$, $u = +58\text{ m}$, $II' = -24\text{ mm} = -24 \times 10^{-3}\text{ m}$, $f = ?$
$\frac{II'}{OO'} = -\frac{v}{u} \implies \frac{-24 \times 10^{-3}}{28} = -\frac{v}{58}$
$v = \frac{24 \times 10^{-3}}{28} \times 58 = 0.049\text{ m}$ (or $49.71\text{ mm}$)
Using lens formula, $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{58} + \frac{1}{0.04971} = \frac{1}{f} \implies f = \mathbf{0.049\text{ m}}$
Q: Calculate the power of the eyelens for viewing object at the longest and shortest distances possible with normal vision, assuming eyelens to retina distance of $2\text{ cm}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 2\text{ cm} = 0.02\text{ m}$, $P = ?$
(i) Longest distance ($u = \infty$):
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{\infty} + \frac{1}{0.02} = 50\text{ m}^{-1}$
$P = \frac{1}{f} = \mathbf{50\text{ D}}$
(ii) Shortest distance ($u = 25\text{ cm} = 0.25\text{ m}$):
$\frac{1}{f} = \frac{1}{0.25} + \frac{1}{0.02} = 4 + 50 = 54\text{ m}^{-1}$
$P = \frac{1}{f} = \mathbf{54\text{ D}}$
Q: A compound microscope consists of two thin converging lenses. The focal length of the objective is $10\text{ mm}$ and that of the eyepiece is $20\text{ mm}$. If an object is placed $11\text{ mm}$ from the objective, the instrument produces an image at infinity. Calculate the separation of the lenses and the magnifying power of the instrument.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $f_o = 10\text{ mm}$, $f_e = 20\text{ mm}$, $u_o = 11\text{ mm}$, $v_e = \infty$
For objective lens: $\frac{1}{f_o} = \frac{1}{u_o} + \frac{1}{v_o} \implies \frac{1}{10} = \frac{1}{11} + \frac{1}{v_o}$
$\frac{1}{v_o} = \frac{1}{10} - \frac{1}{11} = \frac{1}{110} \implies v_o = +110\text{ mm}$
For normal setting ($v_e = \infty$, $u_e = f_e = 20\text{ mm}$):
Separation $L = v_o + u_e = 110 + 20 = \mathbf{130\text{ mm}}$
Magnifying power $MP = \frac{v_o}{u_o} \times \frac{250}{u_e} = \frac{110}{11} \times \frac{250}{20} = \mathbf{125}$
Q: A compound microscope $17\text{ cm}$ long has an eyepiece with a focal length of $2.5\text{ cm}$ and objective with focal length of $0.25\text{ cm}$. What is the magnifying power for normal setting?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $L = 17\text{ cm}$, $f_e = 2.5\text{ cm}$, $f_o = 0.25\text{ cm}$
For normal setting, $u_e = f_e = 2.5\text{ cm}$
$L = v_o + u_e \implies 17 = v_o + 2.5 \implies v_o = 14.5\text{ cm}$
$\frac{1}{f_o} = \frac{1}{u_o} + \frac{1}{v_o} \implies \frac{1}{0.25} = \frac{1}{u_o} + \frac{1}{14.5}$
$\frac{1}{u_o} = \frac{1}{0.25} - \frac{1}{14.5} = \frac{114}{29} \implies u_o = \frac{29}{114}\text{ cm}$
$MP = \frac{v_o}{u_o} \times \frac{25}{u_e} = \frac{14.5}{\frac{29}{114}} \times \frac{25}{2.5} = \mathbf{580}$
Q: A compound microscope consists of two thin lenses, an objective of focal length $20\text{ mm}$ and an eyepiece of focal length $50\text{ mm}$, placed $220\text{ mm}$ apart. If the final image is viewed at infinity, calculate the distance of the object from the objective.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $f_o = 20\text{ mm}$, $f_e = 50\text{ mm}$, $L = 220\text{ mm}$, $v_e = \infty$
For image at infinity: $u_e = f_e = 50\text{ mm}$
$L = v_o + u_e \implies 220 = v_o + 50 \implies v_o = +170\text{ mm}$
Using lens formula for objective: $\frac{1}{f_o} = \frac{1}{u_o} + \frac{1}{v_o}$
$\frac{1}{20} = \frac{1}{u_o} + \frac{1}{170} \implies \frac{1}{u_o} = \frac{1}{20} - \frac{1}{170} = \frac{3}{68}$
$u_o = \frac{68}{3} = \mathbf{22.67\text{ mm}}$
Q: A telescope consists of two converging lenses: an objective of focal length $500\text{ mm}$ and an eyepiece of focal length $50\text{ mm}$. When the telescope is in normal adjustment: (i) What is the separation of the lenses? (ii) Where is the final image located? (iii) Is the image erect or inverted? (iv) What is the magnifying power?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$f_o = 500\text{ mm}$
(i) $L = ?$
(iii) erect or inverted ?
$f_e = 50\text{ mm}$
(ii) $v_e = ?$
(iv) $\text{MP} = ?$
For telescope,
$u_o = \infty$, $\quad v_o = f_o = 500\text{ mm}$
For normal adjustment,
$v_e = \infty$, $\quad u_e = f_e = 50\text{ mm}$
(i) $\begin{aligned}[t] L &= v_o + u_e \\ &= f_o + f_e = 500 + 50 = \mathbf{550\text{ mm}} \end{aligned}$
(ii) The final image is located at infinity.
(iii) The image is inverted ($\because$ it uses two converging lenses).
(iv) $\begin{aligned}[t] \text{MP} &= \frac{v_o}{u_e} \\ &= \frac{f_o}{f_e} = \frac{500}{50} = \mathbf{10} \end{aligned}$
Q: An astronomical telescope has its two-lens space $75.2\text{ cm}$ apart. If the objective lens has a focal length of $70\text{ cm}$, what is the magnifying power of this telescope when viewing with a relaxed eye?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$f_o = 70\text{ cm}$
$L = 75.2\text{ cm}$
$f_e = ?$
$\text{MP} = ?$
For telescope,
$u_o = \infty$, $\quad v_o = f_o$
For relaxed eye,
$v_e = \infty$, $\quad u_e = f_e$
$\begin{aligned}[t] L &= v_o + u_e \\ L &= f_o + f_e \\ 75.2 &= 70 + f_e \\ f_e &= 75.2 - 70 = \mathbf{5.2\text{ cm}} \end{aligned}$
$\begin{aligned}[t] \text{MP} &= \frac{v_o}{u_e} \\ &= \frac{f_o}{f_e} = \frac{70}{5.2} = \mathbf{13.46} \end{aligned}$
Q: A telescope consists of two converging lenses. When in normal adjustment, so that the image of a distance object is formed at infinity, the lenses are $450\text{ mm}$ apart, and the magnifying power is $8$. What are the focal lengths of the two lenses? Is the image erect or inverted?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$f_o = ?$
$L = 450\text{ mm}$
$f_e = ?$
$\text{MP} = 8$
For telescope,
$u_o = \infty$, $\quad v_o = f_o$
For normal adjustment,
$v_e = \infty$, $\quad u_e = f_e$
$\begin{aligned}[t] L &= v_o + u_e \\ 450 &= f_o + f_e \\ f_o &= 450 - f_e \end{aligned}$
$\begin{aligned}[t] \text{MP} &= \frac{v_o}{u_e} \\ 8 &= \frac{f_o}{f_e} \\ 8 &= \frac{450 - f_e}{f_e} \\ 8f_e &= 450 - f_e \\ 9f_e &= 450 \\ f_e &= \mathbf{50\text{ mm}} \\ f_o &= 450 - 50 = \mathbf{400\text{ mm}} \end{aligned}$
The image is **inverted** with respect to the object because this type of telescopes uses two converging lenses.
Q: A telescope consists of two thin converging lenses of focal lengths $100\text{ cm}$ and $10\text{ cm}$ respectively. It is used to view an object $2000\text{ cm}$ from the objective. What is the separation of the lenses if the final image is $25\text{ cm}$ from the eyepiece? Determine the magnifying power for an observer whose eye is close to the eyepiece.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
For telescope, $f_o > f_e$
$f_o = 100\text{ cm}$
$u_o = 2000\text{ cm}$, $v_o \neq f_o$
$L = ?$
$f_e = 10\text{ cm}$
$v_e = -25\text{ cm}$ (virtual image), $u_e \neq f_e$
$\text{MP} = ?$
Using lens formula for objective ($1/f_o = 1/u_o + 1/v_o$):
$\begin{aligned}[t] \frac{1}{100} &= \frac{1}{2000} + \frac{1}{v_o} \\ \frac{1}{v_o} &= \frac{1}{100} - \frac{1}{2000} = \frac{19}{2000} \\ v_o &= \frac{2000}{19} = \mathbf{105.26\text{ cm}} \end{aligned}$
Using lens formula for eyepiece ($1/f_e = 1/u_e + 1/v_e$):
$\begin{aligned}[t] \frac{1}{10} &= \frac{1}{u_e} + \frac{1}{-25} \\ \frac{1}{u_e} &= \frac{1}{10} + \frac{1}{25} = \frac{7}{50} \\ u_e &= \frac{50}{7} = \mathbf{7.14\text{ cm}} \end{aligned}$
$\begin{aligned}[t] L &= v_o + u_e \\ &= 105.26 + 7.14 = \mathbf{112.4\text{ cm}} \end{aligned}$
$\begin{aligned}[t] \text{MP} &= \frac{v_o}{u_e} \\ &= \frac{105.26}{7.14} = \mathbf{14.74} \end{aligned}$
Q: In a laser unit, laser beam of wavelength $6328\text{ \AA}$ is emitted. How many photons are released per second if the output power is $1\text{ mW}$? (Plank constant, $h = 6.62 \times 10^{-34}\text{ Js}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$\lambda = 6328\text{ \AA} = 6328 \times 10^{-10}\text{ m}$
$h = 6.62 \times 10^{-34}\text{ J s}$, $\quad c = 3 \times 10^8\text{ m s}^{-1}$
number of photon per second = $?$
$P = 1\text{ mW} = 1 \times 10^{-3}\text{ W} = 1 \times 10^{-3}\text{ J s}^{-1}$
Output energy in $1\text{ s}$ is $10^{-3}\text{ J}$.
Energy of a photon,
$\begin{aligned}[t] E &= \frac{hc}{\lambda} \\ &= \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6328 \times 10^{-10}} \\ &= \mathbf{3.14 \times 10^{-19}\text{ J}} \end{aligned}$
Number of photons per second
$\begin{aligned}[t] &= \frac{\text{Output energy per second}}{\text{Energy of a photon}} \\ &= \frac{1 \times 10^{-3}}{3.14 \times 10^{-19}} = \mathbf{3.18 \times 10^{15}\text{ photons}} \end{aligned}$
Q: How many photons are emitted in $1\text{ s}$ by a $7.5\text{ mW}$ of $\text{CO}_2$ laser with the wavelength of $10.6\text{ }\mu\text{m}$? (Plank constant, $h = 6.62 \times 10^{-34}\text{ J s}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$\lambda = 10.6\text{ }\mu\text{m} = 10.6 \times 10^{-6}\text{ m}$
$h = 6.62 \times 10^{-34}\text{ J s}$, $\quad c = 3 \times 10^8\text{ m s}^{-1}$
number of photon per second = $?$
$P = 7.5\text{ mW} = 7.5 \times 10^{-3}\text{ J s}^{-1}$
Output energy in $1\text{ s}$ is $7.5 \times 10^{-3}\text{ J}$.
Energy of a photon,
$\begin{aligned}[t] E &= \frac{hc}{\lambda} \\ &= \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10.6 \times 10^{-6}} \\ &= \mathbf{1.87 \times 10^{-20}\text{ J}} \end{aligned}$
Number of photons per second
$\begin{aligned}[t] &= \frac{\text{Output energy per second}}{\text{Energy of a photon}} \\ &= \frac{7.5 \times 10^{-3}}{1.87 \times 10^{-20}} = \mathbf{4 \times 10^{17}\text{ photons}} \end{aligned}$
Q: A He-Ne laser emits photons with energies of $3.14 \times 10^{-19}\text{ J}$. What is the colour of the laser light? (Plank constant, $h = 6.62 \times 10^{-34}\text{ J s}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$E = 3.14 \times 10^{-19}\text{ J}$
$h = 6.62 \times 10^{-34}\text{ J s}$, $\quad c = 3 \times 10^8\text{ m s}^{-1}$
$\lambda = ?$
Energy of a photon,
$\begin{aligned}[t] E &= \frac{hc}{\lambda} \\ 3.14 \times 10^{-19} &= \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{\lambda} \\ \lambda &= \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3.14 \times 10^{-19}} \\ \lambda &= 632.48 \times 10^{-9}\text{ m} \\ &= \mathbf{632.48\text{ nm}} \end{aligned}$
The colour of the laser light is **red**.
Q: An optical fibre used for communications has a core of refractive index $1.55$ which is surrounded by cladding of refractive index $1.45$. Calculate the critical angle of the core-cladding boundary.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
refractive index of core, $n_1 = 1.55$,
refractive index of cladding, $n_2 = 1.45$
$\begin{aligned}[t] \sin i_c &= {}^1n_2 \\ &= \frac{n_2}{n_1} \\ &= \frac{1.45}{1.55} \\ i_c &= \sin^{-1} \left(\frac{1.45}{1.55}\right) = \mathbf{69.31^\circ}\text{ (or) }\mathbf{69^\circ\text{ 18'}} \end{aligned}$
Q: The core of an optical fibre has a dimeter $2\text{ }\mu\text{m}$ and refractive index of $1.63$. The refractive index of the cladding is $1.5$. Determine the maximum angle for which the light rays incident on the end of fibre rod to total internal reflection at the core to cladding interface.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$d = 2\text{ }\mu\text{m} = 2 \times 10^{-6}\text{ m}$
refractive index of core, $n_1 = 1.63$,
refractive index of cladding, $n_2 = 1.5$
$\begin{aligned}[t] \sin i_c &= {}^1n_2 \\ &= \frac{n_2}{n_1} = \frac{1.5}{1.63} \\ i_c &= \sin^{-1}\left(\frac{1.5}{1.63}\right) = \mathbf{66.96^\circ} \end{aligned}$
From figure,
$r + i_c + 90^\circ = 180^\circ$
$r = 180^\circ - 90^\circ - i_c = 90^\circ - 66.96^\circ = \mathbf{23.04^\circ}$
Refraction at air-core interference,
By using Snell's law,
$\begin{aligned}[t] n_1 &= \frac{\sin i}{\sin r} \\ \sin i &= n_1 \sin r \\ &= 1.63 \times \sin 23.04^\circ \\ i &= \sin^{-1}(1.63 \times \sin 23^\circ) = \mathbf{39.64^\circ} \end{aligned}$
Q: A $10\text{ W}$ fluorescent lamp has a luminous intensity of $35\text{ cd}$. Find luminous flux it emits.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$I_v = 35\text{ cd}$, $\quad P = 10\text{ W}$, $\quad \Phi = ?$
Solid angle for a sphere, $\Omega = 4\text{ }\pi\text{ sr}$
$\begin{aligned}[t] I_v &= \frac{\Phi}{\Omega} \\ \Phi &= I_v \Omega \\ &= 35 \times 4\text{ }\pi = \mathbf{439.8\text{ lm}} \end{aligned}$
Q: The luminous intensity of a light source is $500\text{ cd}$. Find the illuminance of a surface which is $10\text{ m}$ from source if light falls normally on it.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$I_v = 500\text{ cd}$, $\quad E_x = ?$
Just below the lamp, $\quad r = 10\text{ m}$, $\quad \theta = 0^\circ$
$\begin{aligned}[t] E_x &= \frac{I_v \cos \theta}{r^2} \\ &= \frac{500 \cos 0^\circ}{10^2} = \mathbf{5\text{ lx}} \end{aligned}$
Q: The luminous intensity of a source is $200\text{ cd}$. The mounting height of the lamp is $6\text{ m}$ from the ground. Find the illuminance on the ground $E_x$: (i) just below the lamp, (ii) $3\text{ m}$ horizontally away from the lamp on the ground.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$I_v = 200\text{ cd}$, $\quad E_x = ?$
(i) Just below the lamp, $\quad r = 6\text{ m}$, $\quad \theta = 0^\circ$
$\begin{aligned}[t] E_x &= \frac{I_v \cos \theta}{r^2} \\ &= \frac{200 \cos 0^\circ}{6^2} = \mathbf{5.55\text{ lx}} \end{aligned}$
(ii) $3\text{ m}$ horizontally away from the lamp on the ground,
$\begin{aligned}[t] r^2 &= 6^2 + 3^2 \\ r &= \sqrt{36 + 9} = \sqrt{45}\text{ m} \\ \cos \theta &= \frac{6}{r} = \frac{6}{\sqrt{45}} = 0.8944 \\ E_x &= \frac{I_v \cos \theta}{r^2} \\ &= \frac{200 \times 0.8944}{(\sqrt{45})^2} \\ &= \frac{200 \times 0.8944}{45} = \mathbf{3.98\text{ lx}} \end{aligned}$
Q: Light from a lamp is falling normally on a surface distant $10\text{ m}$ from the lamp and the illuminance on it is $10\text{ lux}$. In order to increase the illuminance $9$ times, what distance will the lamp be placed from the surface?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Just below the lamp, $\quad \theta = 0^\circ$
$E_{v1} = 10\text{ lux}$, $\quad r_1 = 10\text{ m}$
$E_{v2} = 9 E_{v1} = 90\text{ lux}$, $\quad r_2 = ?$
By inverse square law of illuminance,
$\begin{aligned}[t] E_x &\propto \frac{1}{r^2} \\ E_x r^2 &= \text{Constant} \\ E_{v2} r_2^2 &= E_{v1} r_1^2 \\ 90 r_2^2 &= 10 \times 10^2 \\ r_2^2 &= \frac{10 \times 100}{90} = \frac{100}{9} \\ r_2 &= \frac{10}{3} = \mathbf{3.33\text{ m}} \end{aligned}$
Q: Two light sources of $8\text{ cd}$ and $12\text{ cd}$ are placed on the same side of the photometer screen at a distance of $40\text{ cm}$ from it. Where should a $80\text{ cd}$ source be placed to balance the illuminance?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$I_{v1} = 8\text{ cd}$, $\quad r_1 = 40\text{ cm} = 0.4\text{ m}$
$I_{v2} = 12\text{ cd}$, $\quad r_2 = 40\text{ cm} = 0.4\text{ m}$
$I_{v3} = 80\text{ cd}$, $\quad r_3 = ?$
To balance the illuminance,
$\begin{aligned}[t] E_{v1} + E_{v2} &= E_{v3} \\ \frac{I_{v1}}{r_1^2} + \frac{I_{v2}}{r_2^2} &= \frac{I_{v3}}{r_3^2} \\ \frac{8}{0.4^2} + \frac{12}{0.4^2} &= \frac{80}{r_3^2} \\ \frac{20}{0.4^2} &= \frac{80}{r_3^2} \\ \frac{1}{0.4^2} &= \frac{4}{r_3^2} \\ \frac{1}{0.4} &= \frac{2}{r_3} \\ r_3 &= 0.8\text{ m} = \mathbf{80\text{ cm}} \end{aligned}$